The number of roots of the equation x + 2tanx = π/2 in the interval [0, 2π] is

Q: The number of roots of the equation x + 2tanx = π/2 in the interval [0, 2π] is

(A) 1

(B) 2

(C) 3

(D) infinite

Sol. We have x + 2 tanx = π/2 or tan x = π/4 – x/2.

Now the graphs of the curve y = tanx and y = π/4- x/2, in the interval [0, 2π] intersect at three points.
The abscissa of these three points are the roots of the equation.

Hence (C) is the correct answer.

The number of all possible 5-tuples (a1 , a2 , a3 , a4 , a5 ) such that a1 + a2sinx + a3cosx + a4 sin2x + a5cos2x = 0 holds for all x is

Q: The number of all possible 5-tuples (a1, a2, a3, a4, a5 ) such that
a1 + a2 sinx + a3 cosx + a4 sin2x + a5 cos2x = 0 holds for all x is

(A) zero

(B) 1

(C) 2

(D) infinite

Sol. Since the equation a1 + a2 sinx + a3 cosx + a4 sin2x + a5 cos2x = 0 holds for all values of x ,

a1+ a3 + a5= 0 ( on putting x = 0)

a1- a3+ a5= 0 ( on putting x = π)

⇒ a3 =0 and a1 + a5 = 0 . . . . (1)

Putting x = π/2 and 3π/2 , we get

a1+ a2- a5 = 0 and a1- a2- a5 = 0

⇒ a2 =0 and a1 –a5 =0 . . . . (2)

(1) and (2) give

a1 =a2 = a3 = a5 =0

The given equation reduces to a4sin2x = 0

This is true for all values of x , therefore a4 = 0

Hence a1 =a2 = a3 = a4 =a5 = 0.

Thus the number of 5-tuples is one.

Hence (B) is the correct answer.

If [sin^-1cos^-1 sin^-1 tan^-1 x] = 1 , then x belongs to the interval

Q: If [sin-1cos-1 sin-1 tan-1 x] = 1 , where [.] denotes the greatest integer function, then x belongs to the interval

(A) [ tan sin cos1 , tan sin cos sin1 ]

(B) [ tan sin cos1 , tan sin cos sin1 ]

(C) [1 , -1]

(D) [ sin cos tan1, sin cos sin tan1 ]

Sol. We have [ sin-1 cos-1 sin-1 tan-1x ] = 1

⇒ 1 ≤ sin-1 cos-1 sin-1 tan-1 x ≤ π/2

⇒ sin1 ≤ cos-1 sin-1 tan-1 x ≤ 1

⇒ cos sin1 ≥ sin-1 tan-1 x ≥ cos1

⇒ sin cos sin1 ≥ tan-1 x ≥ sin cos1

⇒  tan sin cos sin1 ≥ x ≥ tan sin cos 1

Hence x ∈ [ tan sin cos1, tan sin cos sin1]

Hence (A) is the correct answer.

The number of points inside or on the circle x^2 + y^2 = 4 satisfying tan^4x + cot^4x + 1 = 3sin^2y is

Q: The number of points inside or on the circle x2 + y2 = 4 satisfying
tan4x + cot4x + 1 = 3sin2y is

(A) one

(B) two

(C) four

(D) infinite

Sol. tan4x + cot4x + 1 = (tan2x – cot2x)2 + 3 ≤ 3

3 sin2y  ≤ 3

⇒ tan2 x = cot2 x , sin2y = 1

⇒ tanx = ± 1, siny = ±1

⇒ x = ± π/4, ± 3π/4 , . . .

But x2 ≤  4 ⇒ -2 ≤  x ≤  2 ⇒ x = ± π/4 only

Siny = ± 1 ⇒  y = ± π/2 ,  3π/2 , . . ..

But y2 ≤ 4  ⇒ y = ± π/2  only.

So four solutions are possible.

Hence (C) is the correct answer.

Indicate the relation which is false

Q: Indicate the relation which is false

(A) tan | tan-1 x | = | x |

(B) cot | cot-1 x | = x

(C) tan-1 | tan x | = | x |

(D) sin | sin-1 x | = | x |

Sol. Since $ \large | tan^{-1}x | = \left\{\begin{array}{ll} tan^{-1}x \; , if \; 0 \leq tan^{-1}x \le \pi/2 \\ -tan^{-1}x \; , if  \; -\pi/2 < tan^{-1}x < 0 \end{array} \right. $

$ \large = \left\{\begin{array}{ll} tan^{-1}x \; , if \; x \ge 0 \\ -tan^{-1}x \; , if x < 0 \end{array} \right. $

⇒ | tan-1x | = tan-1 |x| ∀ x ∈ R

⇒ tan| tan-1x | = tan tan-1 |x| = |x|

Hence (A) is the correct answer.

Likewise sin | sin-1x| = sin sin-1|x| = |x| ∀ |x| ≤ 1

Hence (D) is the correct answer.

| cot-1x | = cot-1x as 0 < | cot-1x| < π ∀ x ∈ R

⇒ cot |cot-1x| = cot cot-1x = x .

Hence (B) is the correct answer.

Since | tanx | is not necessarily always equal to tan|x|

Hence tan-1 | tanx | ≠ | x| .